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How do I calculate the resistor value for a simple LED circuit?

I have a simple circuit: The max rating (current) of the LED is 30mA. How can I work out what the resistance of the resistor needs to be? By using ohm's law I found it to be 3/0.03 = 100 ohm. However using software called Yenka, and trial and error I got the minimum possible resistance to be 36 Ω. However, if I use a 35 Ω resistor, then the LED breaks. Is the software wrong, or (more likely) am I doing something wrong? led resistance share|improve this question at 10:54 Johan 1,5441115 asked May 31 '11 at 10:28 Jonathan. 18219

What's the voltage drop on the diode? Also, here's a nice calculator that worked for me in the past. – AndrejaKo May 31 '11 at 10:34 I am new to electronics, could you say what voltage drop is? – Jonathan. May 31 '11 at 10:36 Maybe the software is assuming that the voltage drop of diode is 1.9V which is used for high-brightness red LEDs. But usually most of the LEDs have 1.7V drop and that would make the total voltage of the circuit 1.3V and minimum resistance at 43ohms. It is always best to use a 47ohm resistor. – Rick_2047 May 31 '11 at 10:42 LED voltages for different colors – endolith May 31 '11 at 14:49 You can also think about the voltage across the LED from the LEDs perspective: When you apply 0V to the LED, there will also be no current through it. When you increase the voltage, almost no current will develop until you reach approximately (!) 1.6V (exact values will vary from LED to LED). A further increase of the voltage will then cause a drastic increase in current. This is exactly the point where you need the resistor: The resistor will define the current above the voltage where the LED starts to conduct heavily. You choose a resistor for the difference between supply and "LED voltage". – zebonaut May 31 '11 at 16:45 feedback

4 Answers

 

You'll have to check the datasheet or measure it to know how much voltage drops over your LED. Let's say this is 2V. Then the voltage over the resistor is the difference between your power supply (3V) - voltage over the LED (2V) = 1V. To get 30 mA through the resistor (and thus also the LED) your resistor has to be 1V / 30 mA = 33 Ohm. If your LED voltage is lower the current will be somewhat higher, but the LED shouldn't break! share|improve this answer edited May 31 '11 at 11:13 answered May 31 '11 at 10:38 stevenvh 82.5k5135247

1   by using a voltmeter in the software the voltage across the LED is 1.9V, I'm guessing that is the voltage drop. – Jonathan. May 31 '11 at 10:47 Could you explain this sentence a bit more: "To get 30 mA through the resistor (and thus also the LED) your resistor has the be 1V / 30 mA = 33 Ohm." Why do we use the difference? Is it that all the voltage needs to be 'used up' before the circuit 'reaches' the battery again. – Jonathan. May 31 '11 at 10:48 @Jonathan - You could use that wording. There's a Kirchhoff's voltage law (KVL) which says that the total voltage of your loop always is zero. That means that the sum of the voltages of your power sources matches the sum of the voltages over your loads. – stevenvh May 31 '11 at 10:55 @Jonathan - More about this. Say you have 2 resistors in series in your circuit, 10 ohm and 20 ohm. That's a total of 30 ohm, so the current will be 3v / 30 ohm = 100 mA. If you calculate the voltages over your resistors you get 100 mA * 10 ohm = 1 V, and 100 mA * 20 ohm = 2V. 1V + 2V = the 3V of your source. It always work! – stevenvh May 31 '11 at 11:00 @stevenh, that explains it well, thank you very much! – Jonathan. May 31 '11 at 11:04 show 5 more comments feedback

up vote 7 down vote

You have to take in count the \$V_{F_{typ}}\$ of the LED. Let's take a red LED as example: \$I_{F_{max}} = 30mA\$ \$V_{F_{typ}} = 1.7V\$ \$V_{F_{max}} = 2.1V\$ \$V_{R_{max}} = 5V\$ The LED and the resistance (R) are in series. The source voltage (3V) is the summation of voltage across the LED (\$V_{F_{typ}}\$) and the voltage across the resistance (\$V_{Res}\$). To calculate R we need to know the voltage on the resistance. Let's use the maximum current, but normally is used 15mA. \$R = \frac{V_{Res}}{I_{F_{max}}}\$ So, \$V_{Res} = V - V_{F_{typ}}\$ \$V_{Res} = 3 - 1.7\$ \$V_{Res} = 1.3V\$ \$R = \frac{V_{Res}}{I_{F_{max}}} = \frac{1,3}{0,03} = 43,33\Omega\$ It's important calculate power dissipation in the resistor, too: \$P_{res} = V_{res} * I_{F_{max}}\$ \$P_{res} = 1.3 * 0,03 = 0,039W\$ (You can use 1/8W or 1/10W) share|improve this answer edited Jun 1 '11 at 10:38 answered May 31 '11 at 11:26 Daniel Grillo 4,56221743

Are you trying to use some kind of math lib on this site or why is you answer formatted with so many $ and {}? – Johan Jun 1 '11 at 6:51 @Johan, I was using MathJaX. They changed the delimiter. Now it's correct. Thanks – Daniel Grillo Jun 1 '11 at 10:40 Now it looks really nice :) – Johan Jun 1 '11 at 13:10 feedback

You have already gotten some direct answers, but here are a few other things to think about. You didn't state it explicitly, but you implied you want 30mA thru the LED. Is this really correct? Certainly there are LEDs and applications for which this is valid. This is not meant as a offence, but if you're here asking a basic Ohm's law question you probably have a common T1 or T1-3/4 LED which is most likely rated for 20mA. Despite what the maximum allowed current is, you need to consider what the purpose of the LED is. Is it for illumination? In that case you probably do want the maximum light you can get. However, your picture implies it's a red LED, so I'm guessing not. If this is just a indicator, you don't need to run it anywhere near its maximum current unless there are unusual circumstances like it needs to be visible in sunlight. Full current can make the result uncomfortably bright as a indicator for a indoor device. I usually run 5mA thru a 20mA LED for indicator use. Is the power coming from a battery? If so, it is likely worth getting a more efficient LED and running it at a lower current. My second comment is about your equation, "3/0.03 = 100 Ohm". Please be more careful with units. That will not only communicate your intentions better, but make it more likely you think about the underlying physics better. Technically your equation is incorrect since the ratio of two dimensionless numbers can't result in a value in Ohms. From context, you meant 3 Volts divided by 0.03 Amps = 100 Ohms. While that would at least make it correct, I recommend to write values in electronics in engineering notation. That means use a power of 1000 such that the value is at least 1 but less than 1000, then apply the appropriate prefix. In this case, 0.03 Amps is better said as 30 milliAmps, or 30 mA. I know this can seem arbitrary and inconvenient to beginners, but it's worth learning and getting used to. After a while you'll form a picture in your mind what a milliamp, microvolt, kiloOhm, etc, is. Those skilled in the art whom you hope will answer your question likely had to look at 0.03 Amps and then think to themselves "Ah, he means 30 mA". share|improve this answer answered Jun 4 '11 at 14:40 Olin Lathrop 65.7k160149

Thank you :) I'm just using software to start with, which said the current through the LED is 30mA, I took this to mean it must be 30mA, thank you for correcting this. Is it ok to put units in an equation? like 3V/30mA = 100 Ohms? I put 0.03 Amps because I wasn't sure what degree of ohms the equation would result in? If I had written the equation in milliamps and pressed 3/30 on a calculator what would be the units of the answer? – Jonathan. Jun 4 '11 at 15:00 feedback

I'm new to this web site and don't yet have a good understanding of answers versus comments. I tried to answer your comment as a comment, but apparently they have a small character limit. It is not only OK to put units in equations, but it is wrong not to. Your "3V / 30mA = 100 Ohms" is exactly how it should be written. You bring up a good issue with a calculator though. Calculators generally don't handle units. They basically operate as if everything is a dimensionless quantity. This is where using the scientific notation capability of your calculator becomes very useful. To compute the resistance you do 3 / 30e-3. A value with a small exponents like 30e-3 might be easy to convert in your head to .030, but consider doing a computation with 22pF. 22e-12 is easy enough to enter, but what's the chance of getting .000000000022 right? share|improve this answer answered Jun 5 '11 at 13:51 Olin Lathrop 65.7k160149

Thanks, I guess you have to know the "degrees" off by heart, if your not using a computer. If a comment is too long you can either edit your answer (generally if it's expanding something already in the answer), or just post more than one comment. – Jonathan. Jun 5 '11 at 14:13 I'm not sure what you mean by "degrees". If you're referring to the power of 1000 multipliers, those are universally used, not limited to electronics. Examples are M (mega, 10**6), k (kilo, 10**3), m (milli, 10**-3), u (micro, 10**-6), etc. – Olin Lathrop Jun 5 '11 at 17:51 yes that is what I'm referring to, sorry I am not clued up on the terminology. I shall look them all up :) Thanks – Jonathan. Jun 5 '11 at 20:11